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Activated Sludge Process Design1Jae K. (Jim) Park, ProfessorDept. of Civil and Environmental EngineeringUniversity of Wisconsin-MadisonInformation Checklist1.Select the type of biological treatment process.2.Conduct a material mass balance and determine expected range of flows (minimum, average, and peak) and loadings (COD, TSS, nutrients, etc.).3.Determine biological kinetic coefficients (lab studies).4.Develop a preliminary site plan, piping layout, and location of collection boxes, return sludge pumps, etc.5.Obtain design criteria.6.Obtain effluent quality criteria (BOD5, TSS, TN and TP).7.Develop data on settling characteristics of the biological solids.8.Obtain list of equipment manufacturers and provide equipment selection guide.2Biological ReactorDesign Criteria1.Design a BNR process with anaerobic, anoxic, and aerobic reactor.2.Anaerobic will receive influent and returned sludge from clarifier. Anoxic will receive effluent from anaerobic and recycle from aerobic.3.The effluent will have BOD5/TSS/TN/NO3--N/NH4+-N/TP of 10/10/10/8/1/1 mg/L or better, respectively. 4.Provide four independent process trains in parallel.5.Anaerobic and anoxic must be deep, square and >1.5 hr.6.Provide equipment for measuring raw wastewater flow, return activated sludge, waste sludge, and air supply.7.Blowers shall be capable of delivering max. air requirements considering the largest single unit out of service.8.Aeration equipment shall provide complete mixing of MLSS and shall be capable of maintaining a min. of 2.0 mg/L DO.3Design Criteria - continued9.Diffusers and piping shall be capable of delivering 150% of the average air requirements.10.The sludge pump and piping for RAS shall be designed to provide capacity up to 150% of average design flow.11.Internal recycle between aeration and anoxic basins will have a capacity 200% of the average design flow.12.All sidestreams from the sludge-handling facilities (thickeners, digesters, and dewatering units) shall be returned to the aeration basin. 13.The sludge wasting shall be achieved from the common collection box containing the effluent MLSS from aeration basins.14.The basin hydraulics shall be checked at peak design flow plus the design return sludge flow when only three basins are in operation.4Design Criteria - continued12.The biological kinetic coefficients and operational parameters shall be determined from laboratory studies.Sludge age, c = 12 daysY= 0.6 mg VSS/mg BOD5; kd = 0.06 day-1YN= 0.2 mg VSS/mg NH4+-N; kd,N = 0.05 day-1 (N=nitrification)MLVSS = 3,000 mg/L; MLVSS/MLSS = 0.8RAS concentration = 10,000 mg TSS/LBOD5 = 0.68 BODLBiological solids are 65% biodegradable (0.65)13.Other conditionsQpeak = 1.321 m3/sec, Qave = 0.44 m3/secBOD5, S0= 250 mg/L, Org-N= 17 mg/L, NH4+-N= 19 mg/L, NO3--N= 0 mg/L, TN= 36 mg/L, TP= 6 mg/LTSS, X0 = 260 mg/L; primary sludge VSS/TSS = 0.74Anaerobic digester solids content = 0.065

Schematic Flow and Piping Arrangementof Secondary Treatment Facility - ExampleFlow metersReturn sludge pumpAirWASWAS pumpsand blowersWASAirFour aerationbasinsPrimarybypasslineJunction-splitter boxSludgereturnCollection box forsludge wastingFour secondaryclarifiersScum boxTo digesterDivision boxEffluentjunction boxChlorinationfacility6Schematic Flow and Piping Arrangementof Secondary Treatment Facility

7Schematic Flow and Piping Arrangement

8Design Strategy1.Conduct a material mass balance2.Calculate of anaerobic zone3.Calculate c, biomass increase, and of anoxic zone4.Calculate c, biomass increase, and of aerobic zone5.Calculate the dimensions of anaerobic, anoxic, and aerobic reactors6.Design the required mixing equipment7.Design aeration system8.Select influent and effluent structure, develop hydraulic profile

9Design CalculationsFirst iteration1.Calculate influent flow (stream 1)Flow = 0.44 m3/sec 86,400 sec/d = 38,016 m3/dBOD5 = 38,016 m3/d250 g/m3kg/1,000 g = 9,504 kg/dTSS = 38,016 m3/d260 g/m3kg/1,000 g = 9,884 kg/dOrg-N= 38,016 m3/d17 g/m3kg/1,000 g = 646 kg/dNH4+-N= 38,016 m3/d19 g/m3kg/1,000 g = 722 kg/d NO3--N= 0 kg/d, TN= 1,369 kg/d, TP= 228 kg/d2.Calculate primary sludge characteristics (stream 3)BOD5 (34% removal) = 9,504 kg/day 0.34 = 3,231 kg/dTSS (63% removal) = 9,884 kg/day 0.63 = 6,227 kg/dOrg-N (30% removal)= 194 kg/d NH4+-N (0.36% removal)= 2.6 kg/d NO3--N= 0 kg/dTN (14.4% removal)= 197 kg/dTP (16% removal)= 37 kg/d Solids concentration = 4.5%Specific gravity = 1.03Sludge flow rate = 6,227 kg/d 103 g/kg 0.045 g/g 1.03 1 g/cm3 106 cm3/m3 = 134 m3/day10Red: Assumed or predetermined design valueDesign Calculations - continued3.Calculate primary treated effluent (stream 4)Q = 38,016 m3/d - 134 m3/d = 37,882 m3/dBOD5 = 9,504 kg/ - 3,231 kg/d = 6,273 kg/day = 6,273 kg/d 37,822 m3/d 1000 g/kg = 166 mg/LTSS = 9,884 kg/d - 6,227 kg/d = 3,657 kg/day = 3,657 kg/d 37,822 m3/d 1000 g/kg = 97 mg/LOrg-N = 452 kg/d 12 mg/L NH4+-N = 720 kg/d 19 mg/LNO3--N= 0 kg/d TN = 1172 kg/d 31 mg/LTP = 192 kg/d 5 mg/L4.The characteristics of streams 2 and 5 are the same as those for streams 1 and 4, respectively.5.Effluent standards (stream 6) were given as design criteria6.Calculate WAS (stream 7)Eff. BOD5 = Eff. sol. BOD5 + BOD5 of eff. biological solids Eff. sol. BOD5= Total BOD5 VSS 1.42 0.68 =10 mg/L10 mg/L0.651.420.68 = 3.7 mg/L111.42 = VSS/COD; COD = BODL = 0.68 BOD5Design Calculations - continuedIncrease in TVSS due to overall BOD5 removal = Y/(1+kdc) (S0 - S1) Q = 0.6/(1+0.0612) (166 3.7) g/m3 37,882 m3/d kg/1000 g = 2,144 kg/d

Increase in TVSS due to nitrification = YN/(1+kd,Nc) (S0,N - S1,N) Q = 0.2/(1+0.0512) (19* 1**) g/m3 37,882 m3/d kg/1000 g = 85 kg/d*NH4+-N in stream 4**Design criteriaIncrease in TSS = (2,144+85) kg/d 0.8 = 2,789 kg/dTSS in WAS = Increase in TSS - TSS lost in the effluent = 2,789 kg/d - [10 g/m3 (37,882 - QWAS***) m3/d kg/1000 g] = 2,410 kg/dMLSS concentration = 3,000 mg/L 0.8 = 3,750 mg/LQWAS= 2,410 kg/d 3,750 mg/L 106 kg/mg 10-3 L/m3= 643 m3/d

***First iteration assumes that QWAS=0

BOD5 in WAS =2,410 kg/d 0.65 g/g 1.42 g/g 0.68 g/g = 1,513 kg/dSol. BOD5 = 3.7 g/m3 643 m3/d kg/1000 g = 2.4 kg/dTotal BOD5 in WAS = 1,513 kg/d + 2.4 kg/d = 1,515 kg/d12Design Calculations - continuedCompute Org-N, NH4+-N, NO3--N, TN in WASOrg-N = 0.122* kg Org-N/kg TVSS 2,410 kg TSS/d 0.8= 235 kg/d NH4+-N = 1** g NH4+-N/m 643 m/d 10-3 kg/g= 0.64 kg/dNO3--N = 8** g NO3--N /m 643 m/d 10-3 kg/g= 5.1 kg/dTN = 235+0.64+5.1= 241 kg/d* Based on chemical formula C60H87O23N12P** Design criteria

Compute TP in WASTP = TP in influent TP in effluent = 192*** kg/d 1 mg P/L 37,882 m3/d kg/1000 g = 154 kg/d*** From stream 4

7.Qr into anaerobic zone (mass balance)10,000 g/m Qr m/d = 3,750 g/m (Q+Qr) m/dQr = 22,729 m/dQr/Q = 22,729 37,882 = 0.6

8.Qrecycle into anoxic zoneTotal NO3--N lost by denitrification = TN influent (stream 4) - TN effluent - TN WAS= 1,172 379 241 = 552 kg/d13552 kg/d = (22,729 m/d + Qrecycle m/d ) 8 g NO3--N /m 10-3 kg/g Qrecycle = 46,271 m/dQrecycle/Q= 46,271 37,882 = 1.22 1.25This design considers a stripping of P from WAS (stream 8). In order to reduce the P buildup in the BNR system, PO43- ions are released under anaerobic condition, thus alum precipitation of PO4-P as AlPO4 and denitrification of NO3--N in WAS are achieved. A summary of how this process affect the whole process performance is shown below:StreamFlowm/dBOD5kg/dTSSkg/dOrg.-Nkg/dNH4+-Nkg/dNO3--Nkg/dTNkg/dTPkg/d7643151524102350.65.12411548648151529212350.60236154Increase in flow rate is caused by addition of alum, increase in TSS is caused by precipitation of AlPO4 and Al(OH)3, biodegradable fraction = 0.54, organic fraction = 0.66. This example only considers the first iteration.14Design Calculations - continued10.Calculate combined sludge (stream 3 + stream 8)Q = 134 m3/d + 648 m3/d = 782 m3/dBOD5 = 3,231 kg/d + 1,515 kg/d = 4,746 kg/dTSS = 6,227 kg/d + 2,921 kg/d = 9,148 kg/d

11.PS and WAS are mixed and blended. Dilution water is needed to maintain hydraulic loading in the thickener (9.8 m/m-d). Final effluent is used as dilution water. It is assumed a solid loading of 46.9 kg/mdThickener area= 9,148 kg/d 46.9 kg/md = 195 mFlow to the thickener= 9.8 m/md 195 m= 1,911 m/dFlow of dilution water= 1,911 m3/d 782 m3/d = 1,129 m/d12.Characteristics of blended sludge (stream 9)BOD5 and TSS in dilution water = 10 mg/L 1,129 m/d = 11 kg/dBOD5 in stream 9 = 3,231 + 1,515 + 11 = 4,757 kg/dTSS in stream 9 = 9,148 + 11 = 9,159 kg/dBiodegradable fraction of TSS = 4,757 kg/d (9,159 kg/d1.420.68) = 0.54TVSS/TSS= (0.74 6,227(stream 3) +0.66 2,921(stream 8) + 0.811) 9,159

= 0.72#5 #14 #515Design Calculations - continued

Schematic Flow and Piping Arrangement

1613.Calculate thickened sludge (stream 10)TSS = 0.85(efficiency) 9,159 kg/day = 7,785 kg/dayQ = 7,785 kg/d 1,000 g/kg 0.06 g/g 1.03 1 g/mL 106 mL/m3 = 126 m3/dBOD5= 4,757 kg/d 0.85 = 4,043 kg/dOrg.-N = (194(stream 3) + 235(stream 8)) 0.85 = 366 kg/dNH4+-N = (2.6(stream 8) + 0.6(stream 8) + 1.1(dilution water) kg/d) 126 m/d 1,911 m/d = 0.3 kg/dNO3--N = 0 kg/d (complete denitrification)TN = 366 + 0 + 0.3= 366 kg/d TP = 82(non-AlPO4(s)) kg/d 0.85 ( 70 kg/d) + 110(AlPO4(s)) kg/d 1(100% efficiency) = 180 kg/d

14.Stream 11 is calculated with respect to both streams 9 and 1017Design Calculations - continued15.Characteristics of anaerobically digested sludge (stream 12 and supernatant)BOD5 and TS in the supernatant = 3,000 and 4,000 mg/L, respectively; TS in digested sludge = 5%; TVS reduced = 52%; BOD5 stabilization = 60%, Org.-N into to NH4+-N = 15%, Org.-N into to soluble Org.-N = 10%, gas production = 0.936 m3/kg TVS reduced (= 15 ft3/lb TVS)

TVSS(stream 10) =7,785 kg/d 0.72(TVSS/TSS stream 9) = 5,605 kg/dTVSS(stabilized) = 5,065 kg/d 0.52= 2,915 kg/dTVSS(remaining) = 5,605 2,915= 2,690 kg/dTSS(remaining) = 7,785 kg/d 2,915 kg/d= 4,870 kg/d

Q(stream 10)= Q(stream 12) + Q(supernatant)126 m/d = TSS(stream 12) (0.05 1,030) + TSS(supernatant) (0.004 1,000)

TSS(supernatant)= 4,870 kg/d - TSS(stream 12)

Solving for TSS(supernatant)and TSS(stream 12)18Design Calculations - continuedTSS(supernatant)= 136 kg/dTSS(stream 12) = 4,734 kg/dQ(supernatant) = 34 m/dQ(stream 12) = 92 m/d

Others components in stream 12BOD5 = 4,043(stream 10) kg/d (1-0.6) 3000 g/m 34 m/d 10-3 kg/g = 1,515 kg/dOrg.-N = 366(stream 10) (1- 0.1- 0.15) (4,734 kg/d 4,870 kg/d) + 366 kg/d 0.1 (92 m/d 126 m/d) = 294 kg/dNH4+-N = 366 kg/d 0.15 + 0.3(stream 10)kg/d (92 m/d 126 m/d) = 40 kg/dNO3--N = 0 kg/d (complete denitrification)TN = 294 + 40 + 0 = 334 kg/dTP = 70 kg/d (1-0.3) (4,734 kg/d 4,870 kg/d) + 366 kg/d 0.1 (92 m/d126 m/d)= 63 kg/d TP = 63(non-AlPO4(s))kg/d + 110(AlPO4(s))kg/d= 173 kg/d

Biodegradable fraction of TSS = 1,515 kg/d (4,734 kg/d1.420.68)= 0.33TVSS/TSS = 2,690 4,870 = 0.5519Design Calculations - continuedDesign Calculations - continuedOthers components in supernatantBOD5 = 3000g/m 34 m/d 10-3 kg/g = 102 kg/dOrg.-N = 366(stream 10) (1- 0.1- 0.15)(136 kg/d4,870 kg/d) + 366 kg/d 0.1 (34 m/d126 m/d)= 18 kg/dNH4+-N =366 kg/d 0.15 + 0.3(stream 10)kg/d (34 m/d126 m/d)= 15 kg/dNO3--N = 0 kg/d (complete denitrification)TN = 18 + 15 + 0 = 33 kg/dTP = 7(non-AlPO4(s))kg/dGas produced = 0.936 m3/kg 2,915 kg/d = 2,728 kg/d

16.Characteristics of dewatered sludgeThe sludge dewatering facility consists of filter presses. It is assumed that the sludge cake has 25% solids; dewatering facility captures 95% solids; specific gravity of sludge cake is 1.06; organic polymers added to the sludge for conditioning are 0.5% of sludge solids; and 80% added chemicals are incorporated into the sludge cake. Total volume of belt wash water and chemical dilution water = 35 L/kg TSS. No BOD5 is added by chemical conditioning.20Design Calculations - continuedFlow rate of belt wash and chem. dilution water = 4,734 kg/d 35 L/kg 10-3 m/L = 166 m/dOthers constituents in belt wash and chem. dilution waterBOD5 = (379(plant effluent)kg/d 37,882 m/d) 166 m/d = 1.7 kg/dTSS = (379(plant effluent)kg/d 37,882 m/d) 166 m/d = 1.7 kg/dOrg.-N = (38(plant effluent)kg/d 37,882 m/d) 166 m/d = 0.2 kg/dNH4+-N = 0.2 kg/dNO3--N = 1.3 kg/dTN = 1.3 + 0.2 + 0.2 = 1.7 kg/dTP = 0.2 kg/d

TSS removed = 4,734 kg/d 0.95 = 4,497 kg/dOrganic polymer removed = 4,734 kg/d 0.005 0.8 = 19 kg/dTotal TSS = 4,497 + 19 = 4,516 kg/dQ sludge cake = 4,516 (0.25 1,060) = 17 m/d

21Others constituents in the sludge cakeBOD5 = 1,515 kg/d 0.95= 1,439 kg/dOrg.-N = 294 kg/d 0.95= 279 kg/dNH4+-N = 40 kg/d (17 m/d92 m/d)= 7.4 kg/dNO3--N = 0 kg/d (complete denitrification)TN = 279 + 7.4 + 0 = 286 kg/dTP = 170 kg/dQ filtrate = 92 m/d + 166 m/d - 17 m/d = 241 m/dTSS in filtrate = 4,734 kg/d + 1.7 kg/d + 4,734 kg/d 0.005(1-0.8) 4,497 kg/d u= 243 kg/dTotal Q stream 16 = 1,785 m/d + 34 m/d + 241 m/d= 2,060 m/dTotal TSS stream 16 = 1,374 kg/d + 136 kg/d + 243 kg/d= 1,753 kg/d22Design Calculations - continuedDesign Calculations - continuedFinal resultsThe previous computational procedure must be repeated for several iteration to obtain a stable value of influent quality (< 1% of fluctuation). The effluent quantity discharged from the plant will be slightly less than the influent flow due to evaporation loss, loss of water in production of digester gases, and moisture contained in the sludge cake.

Next slide shows the characteristic of streams in final iteration of material mass balance analysis23StreamFlowm/dBOD5kg/dTSSkg/dOrg.-Nkg/dNH4+-Nkg/dNO3--Nkg/dTNkg/dTPkg/d138016950498846467220136922823801695049884646722013692283134323162271942.6019737437882627336574527200117219254005072105498558776013332166393033933933939314393397747176028002730.86.02801768752176033852730.8027417692011500396234684.4048221410132425381803980.303982011118787501444704.107413129715965008320440364193133510514019160357.41418151647783048.2031219024See next slide to know stream numbers

Schematic Flow and Piping Arrangement

25StreamFlowm/dBOD5kg/dTSSkg/dOrg.-Nkg/dNH4+-Nkg/dNO3--Nkg/dTNkg/dTPkg/d152558225716360523.6162168937184110556016124The influent to the BNR (stream 5) is characterized as:BOD5 = 180 mg/LOrg.-N = 13.9 mg/LNH4+-N = 19.5 mg/LNO3--N = 0 mg/LTN = 33.4 mg/LTP = 5.4 mg/LTSS = 137 mg/LQ = 40,050 m/dApplying 5~10% as safety factor for the design of BNR BOD5= 200 mg/LOrg.-N= 15 mg/LNH4+-N= 20 mg/LNO3--N= 0 mg/LTN= 35 mg/LTP= 6 mg/LTSS= 150 mg/LQ = 42,000 m/d

26Design Calculations - continuedDimensions of Anaerobic Zone1.A minimum (HRT) of 1.5 h for anaerobic zone is generally used for typical municipal wastewater.

Vanaerobic= 1.5 h 42,000 m/d 24 h/d= 2,625 m

Dimensions of Anoxic Zone1.Calculate design c (SRT) for denitrification

max. specific growth under field condition (max) = max. specific growth rate (max,DN) temp. correction (FT) DO correction (FDO)

Typical values:max,DN= 0.3 d-1FT,DN= 1.08(T-20) (T=15C)FDO,DN= (1-DOmax)(DOmax=0.10 mg/L)

max,DN= 0.3 1.08(T-20) (1-DOmax) = 0.184 d-127Design Calculations - continued2.Calculate minimum c,DN (SRT) for denitrification

min.c,DN 1 (max,DN kd,DN)

Typical values:kd,DN = 0.04 d-1 Endogenous decay coefficientmin.c,DN = 6.9 d

3.Determine design c,DN (SRT) for denitrification

Design c,DN = safety factor (1.5) min.c,DN = 10.4 da conservative value of 12 d is used for the anoxic zone

4.Calculate sludge growth in the anoxic zone pxYobs,DN = YDN (1 + kd,DN design c,DN ) = 0.405 g VSS/g NO3--N

Typical values:YDN = 0.6 g VSS/g NO3--N

28Design Calculations - continuedpx,DN= Yobs (TNinfluent - Org.-Ncells - NO3--Neffluent NH4+-Neffluent)

Org.-Ncells = 0.122(12.2% N in cells )g Org.-N/g VSS Yobs (BOD5,influent BOD5,sol. effluent) = 0.122 0.405 (200 3.7) = 8.4 mg Org.-N/L

Px,DN= 0.405 g VSS/g NO3--N (35 mg/L 8.4 mg/L 8 mg/L 1 mg/L) = 7.1 mg VSS/L

5.Determine design DN (HRT) for denitrification

Design DN= (24 design c,DN px,DN) (fx X)fx: fraction of heterotrophic microorganism that are capable to carry out denitrification = 0.5X: MLVSS = 3,000 mg/L

Design DN = 1.4 h (assume 1.5 h)

Vanoxic= 1.5 h 42,000 m/d 24 h/d = 2,625 m

29Design Calculations - continuedDimensions of Aerobic Zone (based on both BOD5 stabilization and nitrification)

1.Calculate design c,N for nitrification

max,N=max,N temp. correction (FT,N) DO correction (FDO,N) pH correction (FpH,N)

Typical values:max,N = 0.47 d-1FT,N = e0.098 (T-15)(T=15C)FDO,N = DOmin,N(KDO,N + DOmin,N)(DOmin,N=2.0 mg/L & KDO,N= 1.0 mg DO/L)FpH,N = 1 0.833 (7.2 - pHmin,N)(pHmin,N= 7.2) max, N = 0.313 d-130Design Calculations - continued2.Calculate minimum c,N for nitrification

min.c,N 1 (max,N kd,N)

Typical values:kd,N= 0.05 d-1 endogenous decay coefficientmin.c,N = 3.8 d

3.Determine design c,N for nitrification

Design c,N= safety factor for process consideration (1.5) safety factor for kinetic consideration (2.0) min.c,N

Design c,N= 11.4 d

A conservative value of 12 d is used for the aerobic zone

31Design Calculations - continued4.Calculate the NH4+-N concentration in the effluentUN = (kN NH4+-N) (KN + NH4+-N )(design c,N)-1 =YN UN - kd,N kN = max,N YN; KN = 10(0.051 T 1.158)T=15CTypical value: YN = 0.2 g VSS/g NH4+-N

NH4+-N = 0.3 mg/L 5.Calculate BOD5 consumption for denitrificationCH3OH + 1.5O2 CO2 + 2H2O: 1 mol MeOH = 1.5 mol BODLNO3-+1.08CH3OH + H+0.065C5H7O2N+0.47N2+0.76CO2+2.44H2OBODL consumption = 1.08 mol 32 g/mol MeOH/14 g NO3--N 1.5 mol BODL/mol MeOH = 3.7 g BODL/g NO3--NBOD5,denitrif.= (0.68 g BOD5/g BODL) (3.7 g BODL/g NO3--N ) (TNinfluent - Org.-Ncells - NO3--Neffluent NH4+-Neffluent)BOD5,denitrif. = 44.3 mg BOD5/L

32Design Calculations - continued6.Calculate BOD5 consumption for deoxygenation of DOBOD5,DO.= (0.68 g BOD5/g BODL) (1.3 g BODL/DO) (Rr+reclycle DOmax,N - (1 + Rr+reclycle) DOmax,DN)

Rr+reclycle= Qr/Q + Qrecycle/Q, we will see (slide 35) that Rr+reclycle= 0.6 + 1.7= 2.3, although the first approach was = 0.6 + 1.25 = 1.85 (see slides 12 & 13)DOmax,N = 3.0 mg/LDOmax,DN = 0.1 mg/L

BOD5,DO = 5.8 mgBOD5/L

7.Calculate sludge growth caused by BOD5 removal, px, BOD5Yobs, BOD5=YBOD5 (1+kd,BOD5+design c,aerobic) = 0.349 g VSS/g BOD5

Typical values:YBOD5= 0.6 g VSS/g BOD5kd,BOD5= 0.06 d-1

33Design Calculations - continuedpx,BOD5 = Yobs,BOD5(BOD5influent - BOD5,P-release - BOD5,denitrif - BOD5,DO - BOD5,influent)

px,BOD5 = 0.349 (200 0 44.3 5.8 3.7) = 51.0 mg VSS/L

8.Calculate sludge growth caused by nitrification, px, NYobs,N = YN (1 + kd,N + design c,aerobic ) = 0.125 g VSS/g BOD5

Typical values:YN= 0.2 g VSS/g BOD5kd,N= 0.05 d-1

px,N = Yobs,N (TNinfluent - Org.-Ncells - NH4+-Neffluent)

Org.-Ncells = 0.122(12.2% N in cells ) g Org.-N/g VSS px,BOD5

px,N = 3.5 mg VSS/L34Design Calculations - continued8.Calculate the ratio of heterotrophs and nitrifiers in the MLVSSIncrease in TVSS caused by growth of total cell mass in the mixed culturepx,TVSS= px,BOD5 + px,N + px,DN= 61.6 mg VSS/L

Increase in TVSS due to heterotrophspx,het.= px,BOD5 + px,DN= 58.1 mg VSS/L

Fraction of heterotrophs= 58.1 61.6 = 0.94Fraction of autotrophs= 3.5 61.6 = 0.06

9.Determine design BOD5 and design Ndesign BOD5= (24 design c,aerobic px,BOD5) (fhet. X)design N= (24 design c,aerobic px,N) (fauto. X)fhet.: fraction of heterotrophic microorg. = 0.94fauto.: fraction of autotrophic microorg. = 0.06X: MLVSS = 3,000 mg/L35Design Calculations - continueddesign BOD5= 5.2 hdesign N= 5.6 hThe required design N for nitrification governs the design. Use a conservative value of 6 h. Vaerobic= 6 h 42,000 m/d 24 h/d= 10,500 mQuantity of WASIncrease in total MLVSS = px.TVSS g TVSS/m 42,000 m/d 10-3 kg/g= 2,587 kg TVSS/d Increase in total MLSS = Increase in total MLVSS 0.8= 3,234 kg TSS/dIncrease in total MLSS = TSSeffluent + TSSWAS (42,000 m/d QWAS) (10 g TSS/m) + QWAS (3,000 g TSS/m 0.8 g TVSS/g TSS) (10-3 kg/g) = 3,234 kg TSS/d

QWAS= 752 m/d36Design Calculations - continuedQeffluent= 42,000 752= 41,248 m/d

TSSWAS = 752 m/d (3,000 g VSS/m 0.8 g TVSS/g TSS) (10-3 kg/g) = 2,820 kg TSS/dTSSeffluent= (42,000-752) m/d 10 g/m 103 g/kg= 413 kg/d

BOD5,WAS= 2,820 0.65 1.42 0.68= 1,770 kg/d

Since QWAS increased from 747 m/d to 752 m/d, the other components differ just a little bit from the table shown in slide 23 (stream 7). The new values are shown below: Org.-N= 275 mg/LNH4+-N= 0.75 mg/LNO3--N= 6 mg/LTN= 282 mg/LTP= 211 mg/L37Design Calculations - continuedEstimation of the Qr and Qrecycle

Qr= 0.60 42,000 m/d = 25,200 m/d

Loss of NO3--N by denitrification = TNinfluent TNWAS TNeffluent = 35 g TN/m 42,000 m/d 10-3 kg/g 282 kg TN/d - 10 g TN/m 41,248 m/d 10-3 kg/g = 776 kg NO3--N/d (denitrified)

776 kg NO3--N/d = 8 g NO3--N/m (25,200 m/d + Qrecycle) 10-3 kg/g

Qrecycle = 71,800 m/d

Qrecycle/Q = 71,800 42,000= 1.738Design Calculations - continuedOxygen requirement1.Compute theoretical oxygen requirement (ThOR)

ThOR = Carbonaceous BODL + Nitrogenous BODL

Carbonaceous BODL= BODL,influent BODL,sol.effluent BODL,cells BODL,DN + BODL,DO

= (200/0.68) mg BODL/L (3.7/0.68) mg BODL/L 0.65 TSS VSS/0.8TSS 1.42 (px.TVSS) mg BODL/L - (BOD5,denitrif./0.68) mg BODL/L + 1.3 g BODL/g DO [(Rr+recyle 2.0 mg DO/L)- (1+ Rr+recyle) 0.1 mg DO/L]

= (294 5.5 71.1 65.1 + 5.6) mg O2/L= 157.9 mg O2/L

Operating DO in aerobic zoneMax. DO in anoxic zone39Design Calculations - continuedNitrogenousBODL = 4.57 g BODL/g N (TNinfluent - Org.-Ncells - NH4+-Neffluent) = 4.57 (35 0.122(12.2% N in cells) px,TVSS 1) mg O2/L = 121.1 mg O2/L

ThOR = 157.9 mg O2/L + 121.1 mg O2/L = 279 mg O2/L

Total ThOR = 279 mg O2/L 42,000 m/L 10-3 kg/g = 11,718 kg O2/L40Design Calculations - continued2.Compute standard oxygen requirement under field conditions

where Total ThOR = theoretical oxygen required, kg/day;Csw = oxygen solubility in clean water at standard 20C = 9.15 mg/L;Csw = oxygen solubility in clean water at field temperature, mg/L;C = min. DO maintained in the aeration basin, mg/L; = salinity surface tension factor (0.9 for wastewater) (DO saturation wastewater/DO saturation clean water); = oxygen transfer correction factor for wastewater (0.8~0.9) (oxygen transfer wastewater/oxygen transfer clean water);

41Design Calculations - continuedFs = oxygen solubility correction factor for elevation; and

Habs = absolute atm. pressure at elevation sea level (at 500 m, Habs= 9.7 m);Hbasin = static pressure in the aerobic pressure = 5 m (depth water in the basin);EO2 = oxygen transfer efficiency of air diffuser (8%)Csw at 24C = 8.5 mg/L; C = 2.0 mg/L; = 0.75; = 0.9; Csw = 9.15 mg/L; T=24C

42Design Calculations - continued3.Compute the volume of air requiredAir weight = 1.201 kg/m3; O2 = 23.2% in air by weightTheoretical air required under field condition18,707 kg/day (0.232 1.201 kg/m3) = 839,234m3/day air

43Design Calculations - continuedInfluent, baffle walls, and effluent structure of the anaerobic basin1.Select the arrangement of influent structure: a rectangular (1 m wide and 3 m deep) channel constructed along the side of the first chamber. The influent and return sludge enter the channel. The channel has 16 square (20 cm x 20 cm) submerged ports (4 on the bottom and 12 on the front wall) that distribute the influent along the width of the basin. The hydraulic of the system is established at a peak design flow condition (peak design flow + return sludge)

Ave. design flow to BNR= 42,000 m/d= 0.486 m/sPeak design flow to BNR= 1.321 m/sReturn flow= 0.486 0.6= 0.292 m/sRecycle flow= 0.486 1.7= 0.826 m/s

Ave. design flow to each (4) train= (0.486 + 0.292)/4= 0.195 m/sPeak design flow to each (4) train= (1.321 + 0.292)/4= 0.403 m/s

Ave. discharge to each port= 0.195/16= 0.012 m/sPeak. discharge to each port= 0.403/16= 0.025 m/s

44Design Calculations - continued2.Compute the headloss

For influent channelCd = 0.61, A = 0.20 m 0.20 m, Qave = 0.012 m/s, Qpeak = 0.025 m/s

Dzave.= 0.01 mDzpeak= 0.05 m

Provide 2 baffle walls with 53 orifices in each one. The diameter of each orifice is 20 cm For baffle wallsCd=0.61, A= (p/4 (0.20)2 53) m2, Qave= 0.195 m/s, Qpeak= 0.403 m/s

Dzave.= 0.002 m 0 mDzpeak= 0.008 m 0.01 m

45Design Calculations - continued2.Calculate the head over the effluent weir at the third chamber.

Cd = 0.6, L: length of weir = 5.5 m, Qave = 0.195 m/s, Qpeak = 0.403 m/s

Have.= 0.07 mHpeak= 0.12 m

Provide a free fall of 0.15 m at Qave

46Design Calculations - continuedInfluent, baffle walls, and effluent structure of the anoxic basin1.The influent structure of the anoxic chamber consists of an influent channel along the side of anaerobic chamber. It receives weir overflow from the anaerobic chamber and recycle flow. The intermediate walls and effluent structure are similar to that of anaerobic chambers

Ave. design flow to each (4) train = (0.486+ 0.292 + 1.7 0.486)/4 = 0.402 m/sPeak design flow to each (4) train = (1.321+ 0.292 + 1.7 0.486)/4 = 0.610 m/s

Ave. discharge to each port = 0.402/16 = 0.025 m/sPeak. discharge to each port = 0.610/16 = 0.038 m/s

2.Compute the headloss

47Design Calculations - continuedFor influent channelCd = 0.61, A = 0.20 m 0.20 m, Qave = 0.025 m/s, Qpeak = 0.038 m/s

Dzave.= 0.05 mDzpeak= 0.12 m

Provide 2 baffle walls with 53 orifices in each one. The diameter of each orifice is 20 cm

For baffle wallsCd = 0.61, A = (p/4 (0.20)2 53) m2 , Qave = 0.402 m/s, Qpeak= 0.610 m/sDzave.= 0.01 mDzpeak= 0.02 m

2.Calculate the head over the effluent weir at the third chamber.Cd = 0.6, L: length of weir = 5.5 m, Qave = 0.402 m/s, Qpeak = 0.610 m/sHave.= 0.12 mHpeak= 0.16 mProvide a free fall of 0.15 m at Qave48Design Calculations - continuedInfluent, and effluent structure of the aeration basin1.The influent structure consists of a rectangular channel constructed along the entire width of the aeration basin. The channel has a width of 1 m, and water depth at ave. flow condition is 2 m. The influent discharges over the effluent weir of the third anoxic chamber into the channel at one side of the basin and flows toward the other end. The channel has 20 submerged ports (25 cm x 25 cm) along the bottom of the channel.

Ave. design flow to each (4) train = (0.486+ 0.292 + 1.7 0.486)/4 = 0.402 m/sPeak design flow to each (4) train = (1.321+ 0.292 + 1.7 0.486)/4= 0.610 m/s

Ave. discharge to each port= 0.402/20= 0.020 m/sPeak. discharge to each port= 0.610/20= 0.031 m/s

For influent channelCd=0.61, A= 0.25 m 0.25 m, Qave= 0.020 m/s, Qpeak= 0.031 m/s

Dzave.= 0.01 m; Dzpeak= 0.03 m49Design Calculations - continuedEffluent structure1.Select arrangement of effluent structure: effluent weir boxes, 1-m wide effluent trough (launder), 2 m 2 m effluent box with 0.3 m thick common wall in two aeration basins, and 1.0-m diameter outlet sewer. Provide eight effluent weir boxes, each box having an adjustable rectangular weir 0.75 m in length. Provide stop-gates at each weir box for the flexibility to close some openings if a min. head over weirs is desirable under initial flow conditions.

2.Compute head over weir at average design flowQave + Qr = 0.486 m3/sec + 0.292 m3/sec = 0.778 m3/secFlow per basin = 0.778 m3/sec 4 = 0.195 m3/secFlow per weir box = 0.195 m3/sec 8 = 0.024 m3/secC = 0.6; By trial and error, calculate the head; try L = 0.74 m50Design Calculations - continued

L = 0.75 m - 0.2 0.07 m = 0.74 m3.Compute the head over the weir at peak design flow plus recirculation Max. Q = (1.321 m3/sec + 0.292 m3/sec)/4 = 0.403 m3/secMax. Q per weir box = 0.403/8 = 0.05 m3/sec

L = 0.75 m - 0.2 0.11 m = 0.73 m4.Design the effluent trough (launder)L = 17 m - 2.0 m = 15 my2 = 0.44 m (assume); N = 1

51Design Calculations - continued

= 0.027 m3/sec per m length

Allow 16% for friction losses, turbulence, and bend, and add 0.15 m drop to ensure free fallTotal depth of the effluent trough = 0.52 m 1.16 + 0.15 m= 0.75 m

52Design Details

53Design Calculations - continuedHydraulic Profile

54Design Calculations - continuedMixer Power Requirement for Anaerobic and Anoxic Zones

Volume anaerobic zone= 5.5 m 5.5 m 7.25 m= 219.3 m

Panaerobic =Vanaerobic 0.00094 (m)0.3 (MLSS)0.298 = 219.3 .00094 (1.0087)0.3 (3750)0.298 = 2.5 kW

Since both zone have the same volume, provide identical mixers

Design Capacity of Air Supply System

Theoretical air required = 839,234 m air/d (slide 41)Provide 150% of the theoretical value

Total design air = 1,258,851 m/d = 874 m/min 219 m/min-basin55Design Calculations - continuedVol. of air per kg of BOD5 removed = 1,258,851 m air/d [(200-44.3-3.7) g/m 42,000 m/d 10-3 kg/g] = 198 m air/kg BOD5 removed

Vol. of air per kg of wastewater treated = 1,258,851 m air/d 42,000 m/d = 30 m air/ m wastewater treated

Vol. of air per m of aerobic zone = 1,258,851 m air/d 10,500 m = 120 m air/d- m aerobic zone

56Mechanical Aerators

Fixed aerator

Floating aerator

Turbine aerator57Diffusers

58Design Calculations - continuedDesign of diffused aeration system1.Select diffuser tubeProvide Dacron sock diffusers, standard tube dimension 61 cm 7.5 cm (ID), discharging 0.21 m3 standard air per minute per tube (7.4 cfm).2.Calculate the number of diffuser tubes and arrangementTotal # of diffuser tubes = 874 m3/min 0.21 m3/min = 4162 tubesProvide 4320 diffusers (4 basins 15 rows 4 hanger pipes/row 18 diffuser tubes/hanger pipe)# of diffuser tubes per basin = 4320/4 = 1080# of diffuser tubes per row = 1080/15 = 72Provide four knee and swing joint vertical hanger pipes per row# of diffuser tubes per hanger pipe = 72/4 = 183.Calculate the headloss in pipings and diffusers Total headloss in pipe headers = 5~20 cm (2~8 inches) of water Headloss through diffusers = 40~50 cm (16~20 inches) of water. Allowance for clogging is recommended by manufacturers59Design Calculations - continuedDesign range of air velocities in header pipesPipe diameter, cmVelocity, standard air (m/min)2~8360~5008~25500~90025~40900~105040~601050~120060~801200~135080~1501350~1950Friction factor

Headloss in the straight pipe

where L = equivalent length, m; D = pipe diameter, m; hv = velocity head, mm; Q = air flow (m3/min); P = air supply pressure, atm; T = temp. in pipe, K. T = T0(P/P0)0.283; T0 = ambient air temp., K; P0 = ambient barometric pressure, atm; P = air supply pressure, atm.

60Design Calculations - continuedLosses in pipe fittings (elbows, tees, valves, meters, etc.) may be computed using equivalent pipe lengthL = 55.4 C D1.2where L = equivalent length of pipe fitting in meters for pipe diameter of D and C = factor for equivalent length of pipeFittingsC values for equiv. pipe length Gate valve0.25 Long-radius ell or run of standard tee0.33 Medium-radius ell or run of tee reduced in sized 25%0.42 Standard ell or run of tee reduced in size 50%0.67 Angle valve0.90 Tee through side outlet1.33 Globe valve2.00Losses through air filters, blower, silencer, check valves and fittings should be obtained from the equipment manufacturers.Air filter losses13~76 mm (0.5~3.0 inches)Silencer losses Centrifugal blower13~38 mm (0.5~1.5 inches) Positive displacement blower152~216 mm (6.0~8.5 inches)Check valve losses20~203 mm (0.8~8.0 inches)61Headloss Calculations in Air Pipings62Diam.Air flowaVel.bLcfhLLine Descriptioncmm3/minm/minmmm abHoriz. diffuser header (HDH) - 100.21~1.891343.00.0270.20 9 diffuser tubes (DT)ave. 1.05 bcHanger pipe - HDH w/ 18 DT103.7848112.00.0228.28 cdPipe header to one hanger pipe153.782148.00.0230.76 dePipe header to two hanger pipes157.564288.00.0202.64 efPipe header to three hanger pipes1511.346428.00.0195.65 fgPipe header to four hanger pipes1515.1285618.00.01821.41 ghPipe header to two rows of PHs2030.2496310.00.01710.66 hiPipe header to six rows of PHs3590.7294310.00.0155.16 ijPipe header to 10 rows of PHs45151.2095110.00.0143.81 jkPipe header to 14 rows of PHs50211.68107810.00.0134.09 klPipe header to 18 rows of PHs60272.1696310.00.0132.72 lmPipe header to 22 rows of PHs65332.64100210.00.0122.51 mnPipe header to 26 rows of PHs 70393.12102210.00.0122.42 noPipe header to 30 rows of PHs 70453.60117980.00.01225.76 opAir main supplying air from the 80907.201805179.00.011118.28compressor to the aeration basinTotal214.35

a Air flow is based on standard oxygen requirement under field conditions.b Velocity = Air flow/Areac Estimated equivalent length of pipe = straight length + equivalent length of fittings, valves, bends, tees, meters, etc.

Design Details of Aeration Basin Diffusers and Pipings63

Design Details of Aeration Basin Diffusers and Pipings64Design Calculations - continuedDesign of blowerCentrifugal blowers: 50~70 kPa (7~10 psi), > 15 m3/min (5000 cfm); flow rate controlled by throttling the inlet.Rotary positive displacement blowers: < 45 m3/min In this design, use centrifugal blowers without throttling.1.Calculate supply pressure at the blowerTotal losses in pipings (calculated) =214.35 mmLosses in air filter (manufacturers data) =23.74 mmLosses in silencers (centrifugal)=50.00 mmLosses in compressor pipings and valvings used for parallel combinations (check valve, BV, relief valve, piping and connections, etc.)= 210.00 mmSubmergence (water depth above the diffusers)= 4500.00 mmDiffuser losses, fine bubble diffuser tubes (manufacturers data)= 250.00 mmAllowance for clogging of diffusers and misc. headloss under emergency conditions= 525.20 mmTotal= 5803.00 mm65Blowers

Centrifugal blower

Lobe-type positive displacement blower66Design Calculations - continuedAbsolute pressure = (5.80 m + 10.34 m)/10.34 = 1.56 atm1 std atmosphere = 10.34 m of water2.Compute the volume of airVolume of air to be supplied by the blower (2 basins)= 453.6 m3/minTotal air for 4 basins = 907.2 m3/min3.Select number of blowersProvide a total of five centrifugal blowers each of 230 m3/min (8,200 cfm) design capacity arranged in parallel. Each blower shall have a surge point of approx. 50% below the total design flow. The parallel arrangement shall provide the following operational flexibility:Four blowers will meet 150% average air requirements = 920 m3/minThree blowers will meet the ave. design period air reqs = 690 m3/minTwo blowers will meet the ave. initial period air reqts = 460 m3/minFive blowers will provide one unit for standby service.All blowers shall be provided in a blower building. Proper suction and discharge silencers and suitable foundations for compressors shall be provided.67Design Calculations - continued4.Calculate power requirements - estimated from air flow, discharge and inlet pressures, and air temperature under adiabatic conditions

wherePw= power requirement of each blower, kW;w = air mass flow, kg/sec;R = gas constant, 8.314 kJ/k molK;8.41 = constant for air, kg/k mol;T0 = inlet temperature, K;P0 = absolute inlet pressure, atm;P = absolute outlet pressure, atm; ande = efficiency of the machine (usually 70~80%)

68Design Calculations - continuedDesign of waste sludge system1.Select arrangement for waste sludge withdrawalThe MLSS from four aeration basins is discharged into a collection box. The excess solids are pumped from the collection box to the sludge thickener. The amount of waste solids is 2,820 kg/day or 752 m3/day at a solids concentration of 3,750 mg/L (g/m3).2.Select pumps, piping, and pumping cycleProvide two identical constant-speed waste sludge pumps, each pump capable of independent operation and each having a design pumping capacity of 1.5 m3/min (400 gpm). The pumping duration and frequency of operation shall be controlled by the continuous solids monitoring system in the collection box. The pumping operation shall also be controlled by an automatic time-controlled clock. Only one pump shall operate while the other shall serve as a standby unit.Design of spray nozzleTwo nozzles capable of producing a hard, flat spray about 10 L/min at 103 kN/m2. The pump capable of pumping the total flow of all nozzles at the required nozzle pressure.69Starting of a New Plant1.Start treatment with a portion of incoming wastewater (1/3 or 1/4)2.Use only few units of the plant needed to handle the flow (one to two aeration basins and clarifiers)3.Provide sufficient air to maintain a DO in the mixed liquor between 2~4 mg/L.4.Continue returning the entire sludge until MLSS concentration reaches 400~800 mg/L.5.Slowly increase the influent and bring other units in operation.6.The normal plant operation shall be reached in 2~4 weeks.70Common Operating Problems1.Very stable, dark tan foam on aeration tanks may result from too long sludge age. Increase the sludge wasting.2.Thick billows of white sudsy foam on aeration basin may be due to low MLSS. A reduction in sludge wasting may help the problem.3.Different concentrations of MLSS in different aeration basins may be the result of unequal flow distribution to each basin. Valves and gates should be adjusted to equally distribute the influent and return sludge.4.In diffused aeration systems air rising in large bubbles or clumps in some areas is an indication of faulty diffusers. Clean or replace diffusers, check air supply, and install filter ahead of blowers.5.Low pH of MLSS (pH below 6.7) is an indication of nitrification or acid waste reaching the plant. Check the ammonia and nitrate concentrations in effluent as well as the pH of the influent. Proper control measures include decreasing the sludge age by wasting, lime addition, or proper control of influent.6.Dead spots in the aeration basin may be due to plugged diffusers or under aeration. Check air supply and clean or replace diffusers.71

Diffuser Problem72Operation and Maintenance1.Inspect distribution box daily and clean weirs, gates, to remove solids.2. Remove accumulations of debris from inlet channels and gates, and outlet weir each day.3.Keep daily record of DO in the aeration basin, MLSS concentration, SVI, and sludge age. If any unusually high or low values are found, take corrective measures.4.Clean all vertical walls and channels by squeegee on daily basis.5.Hose down and remove wastewater spills without delay.6.Inspect gratings and exposed metal during daily cleanup for signs of corrosion and deterioration of paint.7.Prepare lubrication chart for mechanical equipment according to manufacturers recommendations.8.Drain each aeration basin annually to inspect underwater portions of the concrete structure, pipings, etc. Replace or repair all defective parts. Patch defective concrete and repaint all clean metal surfaces as required.73